# Integrally closed with roots of identity

Let $\lambda_1,...,\lambda_n$ be roots of unity with $n\geq 2$. Assume that $$\frac{1}{n}\sum_{1}^{n}\lambda_i$$

is integral over $\mathbb{Z}$. Show either $\sum_{1}^{n}\lambda_i=0$ or $\lambda_1=\cdots=\lambda_n$.

I only know the basic definition of integral elements, so does there exists a basic proof to the problem?

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Are $\lambda_j$ supposed to be $n$-roots of unity i.e. $\lambda_j^n=1$ for all $j$? – Davide Giraudo May 10 '12 at 20:55
@DavideGiraudo I don't think we have that assumption. – CC_Azusa May 10 '12 at 21:17

Let $\theta = \displaystyle\sum_{i=1}^n \lambda_i$ and $\Lambda = \frac{1}{n}\theta$.
Using the result that $x$ is integral if and only if $\mathbb Z [x]$ is a finitely generated $\mathbb Z$-module, we conclude that $\theta$ is integral, being a sum of integral elements. (This requires the assumption that $\lambda_i \neq \lambda_j$ for some $i \neq j$). Here and throughout, integral means integral over $\mathbb Z$.
Let $f(X)\in\mathbb Z[X]$ be the monic irreducible polynomial satisfied by $\theta$. Then, $f(n\Lambda) = 0$. So $\Lambda$ satisfies the polynomial $f(nX)$. Since $f(X)$ is irreducible, so must be $f(nX)$. Thus, $f(nX)$ is the minimal polynomial satisfied by $\Lambda$ and its not monic if $n>1$. This contradicts the assumption that $\Lambda$ is integral, unless the degree of $f$ is 1, in which case $\theta=0$ as required.
Very smart way to construct the minimal polynomial for $\Lambda$ – CC_Azusa May 10 '12 at 23:29
If any of the $\lambda_i$ are distinct your number lies strictly inside the unit cirle. I think I know what it's conjugates are, they are among similar sums permuting your roots. I'm a little unsure of what the $\lambda_i$ are as per comments. Then you have an alleged integer all of whose conjugates are $< 1$ in absolute value, and it can only be $0$.